Homework Help:
I need someone to check my final answer

1. The problem statement, all variables and given/known data
If I was to work out the sum of all the even numbers between 1000 and 2000, am I correct in saying that there are exactly 600 even numbers?
Therefore, is the final answer 959400?

Could someone please confirm this?
Thank you.

2. Relevant equations
Sum = n/2[(2a+(n-1)d]
where n is the number of terms, a is the first term and d is the difference between each term.

1. The problem statement, all variables and given/known data
If I was to work out the sum of all the even numbers between 1000 and 2000, am I correct in saying that there are exactly 600 even numbers?

Why would you think there are "exactly 600 even numbers" in 999 consective integers?

Therefore, is the final answer 959400?

Could someone please confirm this?
Thank you.

2. Relevant equations
Sum = n/2[(2a+(n-1)d]
where n is the number of terms, a is the first term and d is the difference between each term.

3. The attempt at a solution

a=1000
d=2
n=600

The answer depends upon whether "between 1000 and 2000" means "including 1000 and 2000" or not.

Another very nice formula for the sum of an arithmetic series is
[tex]n\left(\frac{a_1+ a_n}{2}\right)[/tex]
where a1[/sup] and an are the first and last numbers in an arithmetic sequence of n numbers. However, there are a lot more than 600 even numbers between 1000 and 2000!